Question

If dU = C_VdT show that for the reversible expansion of one mole of an ideal gas: deltaS = C_P ln(T₂/T₁) + R ln(V₂/V₁) if C_V is not equal to f(T).

Answer 1

Dear friend your answer is

dS = dQ / T

The change in entropy is then the inverse of the temperature integrated over the change in heat transfer. For gases, there are two possible ways to evaluate the change in entropy. We begin by using the first law of thermodynamics:

dE = dQ - dW

where E is the internal energy and W is the work done by the system. Substituting for the definition of work for a gas.

dQ = dE + p dV

where p is the pressure and V is the volume of the gas. If we use the definition of the enthalpy H of a gas:

H = E + p * V

Then:

dH = dE + p dV + V dp

Substitute into the first law equation:

dQ = dH - V dp - p dV + p dV

dQ = dH - V dp

is an alternate way to present the first law of thermodynamics. For an ideal gas, the equation of state is written:

p * V = R * T

where R is the gas constant. The heat transfer of a gas is equal to the heat capacity times the change in temperature; in differential form:

dQ = C * dT

If we have a constant volume process, the formulation of the first law gives:

dE = dQ = C (constant volume) * dT

Similarly, for a constant pressure process, the formulation of the first law gives:

dH = dQ = C (constant pressure) * dT

If we assume that the heat capacity is constant with temperature, we can use these two equations to define the change in enthalpy and internal energy. If we substitute the value for p from the equation of state, and the definition of dE in the first energy equation, we obtain:

dQ = C (constant volume) * dT + R * T dV / V

Similarly substituting the value of V from the equation of state, and the definition of dH we obtain the alternate form:

dQ = C (constant pressure) * dT - R * T dp / p

Substituting these forms for dQ into the differential form of the entropy equation gives::

dS = C (constant volume) * dT / T + R * dV / V

and

dS = C (constant pressure) * dT / T - R * dp / p

These equations can be integrated from condition "1" to condition "2" to give: T₁-->T₂ & V₁-->V₂

S2 - S1 = Cv * ln ( T2 / T1) + R * ln ( V2 / V1)

and

S2 - S1 = Cp * ln ( T2 / T1) - R * ln ( p2 / p1)

where Cv is the heat capacity at constant volume, Cp is the heat capacity at constant pressure, and ln is the symbol for the logarithmic function.

If we divide both equations by the mass of gas, we can obtain intrinsic, or "specific" forms of both equations:

ds = cv * ln ( T2 / T1) + R * ln ( v2 / v1)

and

ds = cp * ln ( T2 / T1) - R * ln ( p2 / p1)

where cp and cv are the specific heat capacities. Depending on the type of process we encounter, we can now determine the change in entropy for a gas.

Thank you.